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The following record refer to and operation involving reciprocal levelling.
Instrument at Staff reading on A Staff reading on B Remarks
A 1.155 2.595 Dist. AB = 500m RL of A = 525.500
B 0.985 2.415

Find=

1. the true RL of B

2. the combine correction for curvature and refraction

3. the collimation error

4. whether the line of collimation is included upward and downward.


1 Answer
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1. The true RL of B

The difference of level between A and B

=[(0.2955-1.155)+(2.415-0.985)] / 2

=1.435m (fall from A to B) …………………………………………………………………(1)

RL of B = 525.500-1.435= 524.065m

2. the combine correction for 500m curvature and refraction D = (0.0673DD) = (0.06730.50.5) = 0.0168m (-ve)

3. The collimation error

Let us assume that the line of collimation is inclined upward

Collimation error 500m is = e

When the instrument is at A

Correct staff reading at A = 1.155m

Correct staff reading at B= (2.595 – 0.0168 – e)

True difference of level between A and B =(2.595 – 0.0168 –e ) – 1.155

=1.4232-e ………………………………………………………………………….(2)

From (1) and (2)

1.4232 – e = 1.4350

e= -0.0118

collimation error per 100m = -[(0.0185×100) / 500] = -0.0023m

4. the collimation error is assumed positive but the result is – ve. so the assumption is wrong the line of collimation is actually inclined downward.

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